

22012013, 04:30

#16

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Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
Quote:
Originally Posted by nigel1
In post 7 Dockhead, I think your putting forward the same method as SWL, which to me is just fine tuning the traditional method of calculating a CTS.
The final tide vector will be a best guess, as tidal streams over an hour are not linear. But a best guess is probably as good as it gets.
I've got Admiralty Total Tide Running on the laptop, and stepping through a tidal vector at 5 minute intervals shows the data changing at those intervals, so given the right tools, it is feasible to get the data, but its a bit laborious, and if you manage all the calculations, and then the anchor gets fouled,and you have to start the trip an hour later, its back to the drawing board.
My preferred method is a a rough calculation followed by suck it and see

It looks to me like we are all slouching towards enlightenment
I don't think it is at all hard to come up with an approximation of a last partial hour of tide. I don't think that the average tide over the whole passage is, by any stretch of the imagination, the only convenient approximation. A very easy calculation, which can be done analogue, by feel, for a half hour at the end, would be (A+(B*2))/3, where A is the average tide of the last full hour, and B is the average tide of the entire last hour, of which we only want the first half hour (my formula gives an hourly rate).
Of course if you have five minute (!!) tide data, this all becomes child's play. Where did you get that tool? Link?
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22012013, 04:39

#17

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Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
Quote:
Originally Posted by Seaworthy Lass
WOW, do they get it wrong! They call the course line "the ground track" (it only is for constant current over the entire course). .
"If we then draw this line in, it becomes the water track. It does not matter that the water track has not reached the destination, because you can assume that for a short period, conditions will remain similar, and you can continue on the same course. In practice, you would probably be able to see the destination and could approach it by eye."
A least they admit it will not get you to B (so they do not claim the same precision that Dave did about arriving at B). And what happens if D is NOT close? in my example in the last thread it was a couple of nm away. What happens if the current is significantly different for that portion of the journey? What happens if it adverse enough that you cant reach D?
Seems similar.
Yes, it is very clever! It allows simple computation of complex currents varying in direction as well as degree .
The CD line is then moved parallel to CD until it intersects B in the RYA method. The time taken is calculated that way. The CTS is the same (lines are parallel).
My method has the vector for boat speed going STRAIGHT through B (the destination). No fudging needed .
It is only correct (no matter what world you consider it in, real or otherwise) if the average of the current for the first X hours under consideration is exactly the same as the average current for X+1 hours.
This is not the only problem though. The other problem is which D to choose when one before B is is equidistant to the next one produced after B if you bothered to check where D fell after the next lot of current (this wasnt done in the video, nor was it checked at all if D was in fact closer to B after the next lots of current). I know this isn't taught speaking to someone who recently attended an RYA course.
It amazingly does produce big errors. The quote from the RYA instructor was that examples for exams are chosen where D falls close to B (and maybe otherwise as well??). The instructor said:
"Perform a check to ensure that D lies approx less then 30 minutes away from B, some instructors omit this bit, ie to ensure the tidal data remains valid....
If its greater then 30 minutes (or less) redraw the plot using one more or less tide, (note this is rarely done in examples as the test questions are typically picked to ensure it isn't needed."
In other words it isn't taught by the RYA and students aren't alerted to the possibility of problems if D is a long way from B (and it could be several nm)

I was wrong about the consequences of the water track line intersecting the extended rhumb line not at the destination.
It does NOT give a builtin steering error. It WILL give you the correct CTS provided the tide during the partial hour not calculated, is equal to the average tide over the full hours calculated. It will NOT give you time. The steering error won't be that large unless the partial hour is large compared to the total hours calculated (so this method sucks if you're calculating a 2:30 minutes passage) and/or if the tide is greatly different from the average, during the unanalyzed partial hour.
And I guess that's Dave's argument  that you are already as close as you can possibly get given the uncertainty elsewhere, so you just don't screw with that partial hour.
I am starting to feel hesitantly confident that I understand the gist, at least, of the socalled "RYA method".
I do not agree that it is futile to analyze the last partial hour. In many cases of my own concrete personal experience, the calculations were close and accurate enough to need that further step. I don't think it is ridiculously difficult to come up with a much better approximation of the last partial hour.
Can't wait to see Nigel's five minute tide data  wow!
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22012013, 04:49

#18

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Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
Quote:
Originally Posted by Dockhead
All right, I think I am stumbling towards some kind of understanding of what we are calling the "RYA method"  without much right to call it anything, since apparently no one except Dave knows exactly what it involves.
Here is another helpful resource I found: http://www.tiller.co.uk/downloads/in..._fileID=389DA5
So I finally get the geometry, and finally get why the water track does not point to the destination. It is to avoid the recursive adjustment of the passage time in order to get the lines to converge on the destination.
If we did recursively adjust the passage time, we would get the mathematically perfect answer. However, it's not really practical without using a computer.
Yes, that is correct. There are two unknowns in the last hour  CTS and time subjected to the tide. That is the dilemma I have been trying to solve.
My method addresses this problem very elegantly if I may say so myself (no hiding lights under any bushels here LOL).
So this method has us get it close  to the nearest hour. Then we shrink or expand the triangle proportionately until we get the lines to converge on the destination (and we don't even need to do that, unless we want to know time to destination  we already have our CTS in the previous step). The main thing is that the angle of the water track line, which is our CTS, is determined by the proportion between the lengths of the extended rhumb line and the water track, which is a function of the aggregated position of the end of the tidal vector, the distance over ground to the destination, and the length of the water track determined by average STW. So what happens in effect is that the partial hour, which has been causing all the trouble heretofore, is fudged by averaging up or down from the full hour calculation we end up with.
Yep, you have understood it perfectly
This is the problem.
If there's a big difference between the real tide during the last partial hour, and the average tide, then there can be a pretty big error, I guess. I'm not sure whether or not I'm following in Seaworthy Lass' footsteps as I won't understand her method until she publishes her drawings.

Be patient folks, I need exercise book, pen and ruler for that. Promise to do it tonight.
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22012013, 04:50

#19

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Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
Quote:
Originally Posted by Dockhead
Of course if you have five minute (!!) tide data, this all becomes child's play. Where did you get that tool? Link?

Sorry, no link.
Company I work for subscribes to Admiralty Digital Publications, which includes Total Tide. This has world wide tidal height information, and tidal stream data. Basically, its Admiralty Tide Tables and Tidal Stream Atlas's in a digital format.
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22012013, 04:59

#20

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Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
Quote:
Originally Posted by nigel1
I n post 7 Dockhead, I think your putting forward the same method as SWL, which to me is just fine tuning the traditional method of calculating a CTS.
The final tide vector will be a best guess, as tidal streams over an hour are not linear. But a best guess is probably as good as it gets.
I've got Admiralty Total Tide Running on the laptop, and stepping through a tidal vector at 5 minute intervals shows the data changing at those intervals, so given the right tools, it is feasible to get the data, but its a bit laborious, and if you manage all the calculations, and then the anchor gets fouled,and you have to start the trip an hour later, its back to the drawing board.
My preferred method is a a rough calculation followed by suck it and see

Yes, Dockhead did get it right.
My final vector isn't a "best guess", by the way. It is precise if you measured the proportion I propose. It is reasonably accurate if you eyeball the proportion displayed instead of measuring it
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22012013, 05:08

#21

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Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
Quote:
Originally Posted by Seaworthy Lass
. . . . It is reasonably accurate if you eyeball the proportion displayed instead of measuring it

That's exactly what I am calling an "analogue fudge".
What I mean is that, for example, if you are looking at a tidal atlas, and you see the stream at springs is 4.3 and the stream at neaps is 2.7, and you are two days after springs, instead of laboriously running the numbers digitally, you can eyeball it in an analogue manner, imaging proportions instead of digits, and see pretty damned close what the stream is going to be, and in a blink.
Same thing with a stream which is not perpendicular, same thing with partial hours.
It's the right tool for this job because we are after AVERAGES. We do NOT need to get any particular hour exactly right. A series of well done analogue fudges with this kind of data will get you within a hair of the average of a laborious series of laborious digital calculations. This is really important if you're doing a series of sets of hand calculations  it will make all the difference between a usable technique and an impractical one.
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22012013, 05:14

#22

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Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
Quote:
Originally Posted by nigel1
Sorry, no link.
Company I work for subscribes to Admiralty Digital Publications, which includes Total Tide. This has world wide tidal height information, and tidal stream data. Basically, its Admiralty Tide Tables and Tidal Stream Atlas's in a digital format.

I think it's this: Admiralty TotalTide
It's much too expensive for a leisure sailor.
If the tidal stream data is given with a resolution of five minutes, then it is much higher resolution than the published tide tables.
I have a full set of the Admiralty Tidal Stream Atlases for the English Channel, and they are pretty low resolution  hourly streams and widely spaced diamonds.
If we had access to five minute stream data, we could test another fascinating hypothesis which comes out of this discussion:
"How accurately can we calculate CTS using imprecise hourly tidal data and an analoguefudging method?"
We could compare a really precise calculation using 5minute tide data, with calculations using our own various methods, including the classical RYA method.
That would be really interesting. I will bet a bottle that you don't need the 5minute stream date to come within a couple of degrees.
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22012013, 05:21

#23

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Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
Quote:
Originally Posted by Dockhead
That's exactly what I am calling an "analogue fudge".
What I mean is that, for example, if you are looking at a tidal atlas, and you see the stream at springs is 4.3 and the stream at neaps is 2.7, and you are two days after springs, instead of laboriously running the numbers digitally, you can eyeball it in an analogue manner, imaging proportions instead of digits, and see pretty damned close what the stream is going to be, and in a blink.
Same thing with a stream which is not perpendicular, same thing with partial hours.
It's the right tool for this job because we are after AVERAGES. We do NOT need to get any particular hour exactly right. A series of well done analogue fudges with this kind of data will get you within a hair of the average of a laborious series of laborious digital calculations. This is really important if you're doing a series of sets of hand calculations  it will make all the difference between a usable technique and an impractical one.

OK, dont eyeball my method, measure it with a ruler or dividers (or callipers LOL) and use a calculator and it will be exact
Fudging is optional .
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22012013, 07:46

#24

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Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
Quote:
Originally Posted by Dockhead
It looks to me like we are all slouching towards enlightenment
I don't think it is at all hard to come up with an approximation of a last partial hour of tide. I don't think that the average tide over the whole passage is, by any stretch of the imagination, the only convenient approximation. A very easy calculation, which can be done analogue, by feel, for a half hour at the end, would be (A+(B*2))/3, where A is the average tide of the last full hour, and B is the average tide of the entire last hour, of which we only want the first half hour (my formula gives an hourly rate).
Of course if you have five minute (!!) tide data, this all becomes child's play. Where did you get that tool? Link?

Did you look at my solution (at the end of the last thread)?
As per what we can assume is the RYA method  where the CTS vector intercepts the rhumb line, you calculate the Speed Made Good. Using this SMG calculate time to destination. Using this time amend the tidal vector(s) and CTS vector. It should arrive at destination with correct CTS. If not (either it falls short or beyond destination) then recalculate CMG, follow all the above steps a second time and you should be just about bang on.
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22012013, 08:08

#25

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Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
Quote:
Originally Posted by Lodesman
Did you look at my solution (at the end of the last thread)?
As per what we can assume is the RYA method  where the CTS vector intercepts the rhumb line, you calculate the Speed Made Good. Using this SMG calculate time to destination. Using this time amend the tidal vector(s) and CTS vector. It should arrive at destination with correct CTS. If not (either it falls short or beyond destination) then recalculate CMG, follow all the above steps a second time and you should be just about bang on.

Did you mean this:
"I may have misinterpreted your method from the example, but assuming I understood your method, it seems accurate but a bit labourious. Where your ownship vector (CTS/SRO) crosses the CMG vector (rhumb line) you can calculate speed made good  in the example 11 nm over 3 hours for SMG of 3.67 kts  at that SMG 10 miles is coverered in 2h44m. Shorten the final tidal vector (0.5 kts for 44 mins = 0.367NM). The resultant vector from the end of the truncated tidal vector to the destination should be the course to steer and the length should equal S*T (4 kts * 2h44m = 10.93). I suppose it's possible (particularly with strong currents and/or slow boat speeds) it may not intercept exactly, but would refine the SMG based on where it crosses the rhumb line; in this case you would just do another correction as above to get the final accurate solution."
Well, our methods are all quite different. None of us, neither RYA or the rest of us, calculates or uses VMG. We use aggregated tidal vector to estimate the effect on bearing and distance through water to the destination. So we get a single CTS to the destination, which is the most efficient path (straight path through water). The only difference between RYA and the rest of us is that RYA suggests not calculating the final partial hour, and just using the average vector of the rest of the passage  "inflating" or "deflating" the vector triangle. Several of the rest of us disagree with this  we think that the final partial hour is important, can be calculated, and should be calculated.
This problem had exactly perpendicular tides, so greatly simplified calculation  you can just solve a right triangle, with the rhumb line as "c" and the tidal vector as "b". The water track is almost exactly 10 miles and so time on passage 2.5 hours given boat speed of 4 knots. Three people independently calculated the exact same results, using different methods. These results are quite different from yours; perhaps you have a course change at the end? We are all steering a single CTS.
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22012013, 08:20

#26

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Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
I dont think its so much as "inaccurate" teaching, the RYA are just passing on the basics. There is no difference in their method as in the method used by the Royal and Merchant Navy in the UK, and no doubt, by other maritime academies and schools throughout the world. As "navigators" we should advance our learning and understanding using the basic's as taught to us.
I was taught the basic concepts of celestial navigation at Merchant Navy Colleges, I then went on to learn a shed load more on the subject of my own back as it interested me, and 30 years ago, it was pretty much the only means we had to find ourselves.
I think thrashing the subject out on the forum has been of a great benefit, and I'm pretty sure that the RYA would agree that advancing the vectors so they ended up as close to the destination as possible is a more accurate solution. I guess they also hope that their candidates would suss this out for themselves.
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22012013, 08:27

#27

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Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
Quote:
Originally Posted by Dockhead
Did you mean this:
"I may have misinterpreted your method from the example, but assuming I understood your method, it seems accurate but a bit labourious. Where your ownship vector (CTS/SRO) crosses the CMG vector (rhumb line) you can calculate speed made good  in the example 11 nm over 3 hours for SMG of 3.67 kts  at that SMG 10 miles is coverered in 2h44m. Shorten the final tidal vector (0.5 kts for 44 mins = 0.367NM). The resultant vector from the end of the truncated tidal vector to the destination should be the course to steer and the length should equal S*T (4 kts * 2h44m = 10.93). I suppose it's possible (particularly with strong currents and/or slow boat speeds) it may not intercept exactly, but would refine the SMG based on where it crosses the rhumb line; in this case you would just do another correction as above to get the final accurate solution."
Well, our methods are all quite different. None of us, neither RYA or the rest of us, calculates or uses VMG. We use aggregated tidal vector to estimate the effect on bearing and distance through water to the destination. So we get a single CTS to the destination, which is the most efficient path (straight path through water). The only difference between RYA and the rest of us is that RYA suggests not calculating the final partial hour, and just using the average vector of the rest of the passage  "inflating" or "deflating" the vector triangle. Several of the rest of us disagree with this  we think that the final partial hour is important, can be calculated, and should be calculated.
This problem had exactly perpendicular tides, so greatly simplified calculation  you can just solve a right triangle, with the rhumb line as "c" and the tidal vector as "b". The water track is almost exactly 10 miles and so time on passage 2.5 hours given boat speed of 4 knots. Three people independently calculated the exact same results, using different methods. These results are quite different from yours; perhaps you have a course change at the end? We are all steering a single CTS.

That's the one I meant, but it references the first problem by SWL  the one where the answer was 064º. My method doesn't have a course change at the end  I solve for the final partial hour, making the last tidal vector X minutes in length to give a single CTS. It uses basic pilotage techniques, you don't need to eyeball anything or work out proportions. Worst case you have to do an initial rough calc, then recalc a final solution, but in most cases you could solve it in one.
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22012013, 09:19

#28

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Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
None of these methods are mathematically correct, but they are not meant to be. They are meant to be easy to use and near enough to be useful.
The first mathematical problem is that we're summing discrete data when we should be integrating over continuous data. That's an intractable problem, so the best that we can do is use more granular discrete data, with the aid of computers, which is how NASA navigate in space.
The second problem is that we're projecting a 3dimensional problem onto a 2dimensional plane. That's fine for short distances, but long distances will be seriously wrong.
On top of these mathematical inaccuracies, we have other inaccuracies, such as rough estimations of leeway, tidal variations between our location and standard port, imprecise boat speed, etc.
The RYA philosophy is KISS, keep it simple because a complex approach is likely to introduce errors. I can do 3dimensional trigonometry, but I never navigate that way, not because I'm lazy, but because I would be more likely to introduce a serious error.
RYA stress that it's important to check one's position using other methods (bearings to charted landmarks, depth, GPS, etc.) whenever practicable.
I agree that Seaworthy Lass's method adds some precision, but it is still far from precise and it's debatable whether the added complexity is worth the added precision. For those who understand the maths, this discussion is useful but, for those who do not, I don't think it's anything to worry much about.
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22012013, 09:26

#29

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Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
Quote:
Originally Posted by Lodesman
That's the one I meant, but it references the first problem by SWL  the one where the answer was 064º. My method doesn't have a course change at the end  I solve for the final partial hour, making the last tidal vector X minutes in length to give a single CTS. It uses basic pilotage techniques, you don't need to eyeball anything or work out proportions. Worst case you have to do an initial rough calc, then recalc a final solution, but in most cases you could solve it in one.

The correct answer seems to be 58.3 degrees, and I think it is a mathematically perfect answer, based on the assumptions, ignoring only the fact that the sea surface is not a perfect plane, something absolutely reasonable to assume over distances like these.
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22012013, 09:33

#30

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Re: Inaccurate RYA Teaching : CTS  Quest For a New Method
Quote:
Originally Posted by mcarling
None of these methods are mathematically correct, but they are not meant to be. They are meant to be easy to use and near enough to be useful.
The first mathematical problem is that we're summing discrete data when we should be integrating over continuous data. That's an intractable problem, so the best that we can do is use more granular discrete data, with the aid of computers, which is how NASA navigate in space.

Other than the curvature of the earth problem, immaterial at such distances, I don't think I can agree with this. My solution to SWL's problem, which is admittedly artificial, is mathematically perfect but for the curvature of the earth. That problems lends itself to a simple  and perfect  solution because the tides were all perpendicular to the rhumb line.
SWL's method is also mathematically perfect, within, of course, the limits of the accuracy of the data. But given the fact that tide data is given as an average during a certain time frame of reference (usually an hour), there is no integration problem. Averages are exactly what we need.
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